TITLE: Find $( \dotsb ((2017 \diamond 2016) \diamond 2015) \diamond \dotsb \diamond 2) \diamond 1$ given ...
QUESTION [6 upvotes]: For positive real numbers $a$ and $b,$ let
  $$a \diamond b = \frac{\sqrt{a^2 + 4ab + b^2 - 2a - 2b + 9}}{ab + 6}.$$Find
  $$( \dotsb ((2017 \diamond 2016) \diamond 2015) \diamond \dotsb \diamond 2) \diamond 1.$$


I can't find any quick way to do this. Can anyone help? Thanks in advance!

REPLY [16 votes]: Notice for all positive number $a$, we have
$$a \diamond 2 = \frac{\sqrt{a^2 + 8a + 4 - 2a - 4 + 9}}{2a+6} = \frac{\sqrt{a^2+6a+9}}{2a+6} = \frac12$$
The mess at hand equals to
$$( (\cdots) \diamond 2) \diamond 1 = \frac12 \diamond 1 = \frac{\sqrt{37}}{13}$$